What's RIGHT?

Assume you have a RIGHT circular cylinder of radius R and height H. If H^3 is BIG, AREA is PI*R^2, and H*R is BY, what's the numerical value of RIGHT?

- Carl G.

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The alphametic PI*R^2 = AREA has a unique solution A=4,R=7,E=0,P=9,I=6. The alphametic H^3 = BIG has a unique solution H=8,B=5,I=1,G=2. Then H*R=56 so Y=6 (oops, Y=I... why not leave out the H*R=BY?). Now RIGH = 7628, but what's T? There is only one unassigned digit, namely 3, so that must be it:

RIGHT = 76283.

Robert Israel This e-mail address is being protected from spam bots, you need JavaScript enabled to view it Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada

Big oops. I quickly wrote the puzzle during my lunch break and was checking off which digits were still available. I somehow failed to check off that the letter I was 6. The letter I can't be both 6 and 1.

Carl G.

Yeah, I got the same inconsistency with the value of 'I', so I'm not sure that the problem has a valid answer.

I tried this allowing leading zeroes. This gives: H^3=BIG has one more solution: H=3, B=0, I=2, G=7

Then, H*R=BY becomes: 3*R=Y R cannot be 0,2, or 3- since they have been used already, to R=1. But then Y=3=H. Thus, there are no solutions, even allowing leading zeroes.

I assume that no digits can be repeated, otherwise the trivial solution (all digits are 0) works.